Hydrogen gas is contained in a closed vessel at 1 atm (100 kPa) and 300 K. (a) Calculate the mean speed of the molecules. (b) Suppose the molecules strike the wall with this speed making an average angle of 45° with it. How many molecules strike each square metre of the wall per second?

Use R = 8.31 JK^{-1} mol^{-1}

#### Solution

Here,

P = 10^{5} Pa

T = 300 K

For H_{2}

M = 2×10^{-3} kg

(a) Mean speed is given by

\[ < v > = \sqrt{\frac{8RT}{\pi M}}\]

\[= \sqrt{\frac{8 \times 8.3 \times 300 \times 7}{2 \times {10}^{-3} \times 22}}\]

\[ = 1780 {\text { ms}}^{-1} \]

Let us consider a cubic volume of 1m^{3}.

V = 1 m^{3}

Momentum of 1 molecule normal to the striking surface before collision = mu sin 45°

Momentum of 1 molecule normal to the striking surface after collision = - mu sin 45°

Change in momentum of the molecule = 2mu sin 45° = \[\sqrt{2}mu \]

Change in momentum of n molecules = 2mnu sin 45° = \[\sqrt{2}mnu\]

Let Δt be the time taken in changing the momentum.

Force per unit area due to one molecule \[= \frac{\sqrt{2}mu}{\Delta t} = \frac{\sqrt{2}mu}{\Delta t}\]

Observed pressure due to collision by n molecules\[ =\frac{\sqrt{2}mnu}{\Delta t} = {10}^5\]

\[ n = \frac{\frac{\sqrt{2}mnu}{\Delta t}}{\frac{\sqrt{2}mu}{\Delta t}} = \frac{{10}^5}{\sqrt{2}mu}\]

\[ 6.0 \times {10}^{23} \text { molecules } = 2 \times {10}^{-3} kg\]

\[ 1 \text { molecule } =\frac{2 \times {10}^-{3}}{6 \times {10}^{23}} = 3.3 \times {10}^{-27} kg \]

\[\Rightarrow n = \frac{{10}^5}{\sqrt{2} \times 3.3 \times {10}^{-27} \times 1780} = 1.2 \times {10}^{28}\]